NCERT Exemplar Problems Class 8 Mathematics Exponents and Powers

NCERT Exemplar Problems Class 8 Mathematics  Chapter 8 Exponents and Powers

Multiple Choice Questions
Question. 1 In 2ⁿ, n is known as
(a) base (b) constant
(c) exponent (d) variable
Solution.
(c) We know that an is called the nth power of a; and is also read as a raised to the power n.
The rational number a is called the base and n is called the exponent (power or index). In the same way in 2ⁿ,n is known as exponent.

Question. 2 For a fixed base, if the exponent decreases by 1, the number becomes
(a) one-tenth of the previous number
(b) ten times of the previous number
(c) hundredth of the previous number
(d) hundred times of the previous number
Solution.
(a) For a fixed base, if the exponent decreases by 1, the number becomes one-tenth of the previous number.

Question. 3

Solution.

Question. 4 The value of 1/4^-2 is
Solution.

Question 5 The value of 3⁵   ÷ 3^-6  is
(a) 3⁵     (b) 3^-6        (c) 3¹¹       (d) 3^-11
Solution.

Question. 6

Solution.

Question. 7

Solution.

Question. 8 The multiplicative inverse of 10^-100 is
(a) 10      (b) 100     (c) 10¹⁰⁰    (d)10^-100 
Solution.

Question.9 The value of (-2)^2×3-1 is
(a) 32      (b) 64       (c) -32    (d) -64
Solution.

Question.10

Solution.

Question. 11

Solution.

Question. 12 If x be any non-zero integer and w, n be negative integers, then x^m  x xⁿ is equal to
(a) x^m     (b)x^(m+n)  (c) xⁿ        (d) x^(m-n)
Solution.

Question. 13 If y be any non-zero integer, then y⁰ is equal to
(a) 1                (b) 0                (c) – 1               (d) not defined
Solution.

Question.14 If x be any non-zero integer, then _X^-1 is equal to
(a) x           (b) 1/x           (c) – x         (d) -1/x
Solution.

Question. 15

Solution.

Question. 16

Solution.

Question. 17

Solution.

Question. 18

Solution.

Question. 19

Solution.

Question. 20

Solution.

Question. 21

Solution.

Question. 22

Solution.

Question. 23

Solution.

Question.24 The standard form for 0.000064 is
(a) 64 x 10⁴   (b) 64 x 10^-4  (c) 6.4 x 10⁵ (d) 6.4 x 10^-5
Solution.
(d) Given, 0.000064 = 0. 64 x 10^-4 =6.4 x 10^-5
Hence, standard form of 0.000064 is6.4 x 10^-5.

Question. 25 The standard form for 234000000 is
(a) 2.34 x 10⁸   (b) 0.234 x 10⁹   
(c) 2.34 x 10^-8   (d) 0.234 x 10^{– 9}   
Solution.
(a) Given, 234000000 = 234 x 10⁶ = 2.34 x 10⁺⁶  = 2.34 x 10⁸
Hence, standard form of 234000000is2.34 x 10⁸.

Question.26 The usual form for 2.03 x 10^{-5 is}    
(a) 0.203 (b) 0.00203 (c) 203000 (d) 0.0000203
Solution.

Question. 27

Solution.

Question. 28

Solution.

Question. 29

Solution.

Question. 30

Solution.

Question. 31

Solution.

Question. 32

Solution.

Question. 33

Solution.

Fill in the Blanks
In questions 34 to 65, fill in the blanks to make the statements true.

Question. 34 The multiplicative inverse of 10¹⁰ is_________
Solution.

Question.35 a³ x a^-10= _________
Solution.

Question.36 5⁰ = _________
Solution.

Question.37 5⁵ x 5^-5= _________
Solution.

Question.38

Solution.

Question. 39 The expression for 8^-2 as a power with the base 2 is_________
Solution.

Question. 40 Very small numbers can be expressed in standard form by using_________
exponents
Solution.
Very small numbers can be expressed in standard form by using negative exponents, i.e. 0.000023 = 2.3 x 10^-3

Question. 41 Very large numbers can be expressed in standard form by using
exponents.
Solution.
Very large numbers can be expressed in standard form by using positive exponents,
i.e. 23000 = 23 x 10³ =2.3 x 10³ x 10¹ =2.3 x 10⁴

Question. 42 By multiplying (10)⁵ by (10)^-10, we get
Solution.

Question.43

Solution.

Question.44

Solution.

Question.45

Solution.

Question.46

Solution.

Question.47

Solution.

Question.48

Solution.

Question.49

Solution.

Question.50

Solution.

Question.51

Solution.

Question.52

Solution.

Question.53 The value of 3 x 10^-7   is equal to_______
Solution
Given, 3 x 10^-7   = 3.0 x 10^-7   
Now, placing decimal seven place towards left of original position, we get 0.0000003. Hence, the value of 3 x 10^-7 is equal to 0.0000003.

Question.54 To add the numbers given in standard form, we first convert them into number with_______exponents.
Solution.
To add the numbers given in standard form, we first convert them into numbers with equal exponents.
e.g. 2.46 x 10⁶   + 24.6 x 105 = 2.46 x 10⁵ + 2.46 x 10⁶ = 4.92 x 10⁶

Question.55 The standard form for 32500000000 is_______.
Solution.
For standard form, 32500000000 = 3250 x 10² x 10² x 10³
= 3250 x 10⁷ = 3.250 x 10¹⁰ or 3.25 x 10¹⁰
Hence, the standard form for 32500000000 is 3.25 x 10¹⁰.

Question. 56 The standard form for 0.000000008 is_______.
Solution.
For standard form, 0.000000008 = 0.8 x 10^-8= 8 x  10^-9 =8.0 x 10^-9
Hence, the standard form for 0.000000008 is 8.0 x 10-9

Question.57 The usual form for 2.3 x 10^-10 is_______.
Solution. For usual form, 2.3 x 10^-10 = 0.23 x 10^-11
= 0.00000000023
Hence, the usual form for 2.3 x 10^-10 is 0.00000000023.

Question. 58 On dividing 8⁵ by_______. we get 8.
Solution.

Question. 59

Solution.

Question. 60

Solution.

Question.61

Solution.

Question.62

Solution.

Question.63

Solution.

Question.64

Solution.

Question.65

Solution.

True / False
In questions 66 to 90, state whether the given statements are True or False.

Question.66

Solution.

Question.67

Solution.

Question.68

Solution.

Question.69 24.58 = 2 x 10 + 4 x 1+5 x 10 + 8 x 100
Solution. False
R H S = 2 x 10+ 4 x 1+ 5 x 10+ 8 x 100=20+ 4 + 50+ 800=874 L H S ≠ R H S

Question.70 329.25 = 3 x 10²+ 2 x 10¹ + 9 x 10⁰ + 2 x 10^-1 + 5 x 10^-2
Solution.

Question.71

Solution.

Question.72

Solution.

Question.73

Solution.

Question. 74 5° = 5
Solution.

Question. 75 (-2)° = 2
Solution.

Question.76

Solution.

Question. 77 (-6)° = – 1
Solution.

Question. 78

Solution.

Question. 79

Solution.

Question. 80

Solution.

Question. 81

Solution.

Question. 82

Solution.

Question. 83

Solution.

Question. 84

Solution.

Question.85 The standard form for 0.000037 is 3.7 x 10^-5
Solution. True
For standard form, 0.000037 = 0.37 x 10^-4= 3.7 x 10^-5

Question. 86 The standard form for 203000 is 2.03 x 105.
Solution. True
For standard form, 203000 = 203 x 10 x 10 x 10 = 203 x 10³
= 2.03 x 10² x 10³= 2.03 x 10⁵

Question. 87 The usual form for 2 x 10^-2 is not equal to 0.02.
Solution.

Question. 88 The value of 5^-2 is equal to 25.
Solution. False

Question. 89 Large numbers can be expressed in the standard form by using positive exponents.
Solution.True
e.g. 2360000 = 236 x 10 x 10 x 10 x 10= 236 x 10⁴
‘ = 2.36 x 10⁴ x 10²=2.36 x 10⁶

Question. 90

Solution.

Question. 91 Solve the following,

Solution.

Question. 92 Express 3^-5 x 3^-4 as a power of 3 with positive exponent.
Solution.

Question. 93 Express 16^-2 as a power with the base 2.
Solution.

Question. 94

Solution.

Question. 95

Solution.

Question. 96 Express as a power of a rational number with negative exponent.

Solution.

Question. 97 Find the product of the cube of (-2) and the square of (+4).
Solution.

Question.98 Simplify

Solution.

Question. 99 Find the value of x, so that

Solution.

Question 100 Divide 293 by 1000000 and express the result in standard form.
Solution.

Question. 101

Solution.

Q. 102 By what number should we multiply (-29)°, so that the product becomes (+29)°.
Solution.

Question. 103 By what number should (-15)^-1 be divided so that quotient may be equal to (-15)^-1 ?
Solution.

Question.104 Find the multiplicative inverse of (-7)²÷ (90)^-1
Solution.

Question.105

Solution.

Question.106 Write 390000000 in the standard form.
Solution.

Question. 107 Write 0.000005678 in the standard form.
Solution.
For standard form, 0.000005678 = 0.5678 x 10^-5= 5.678 x 10^-5 x 10^-1= 5.678 x 10^-6 Hence, 5.678 x 10^-6 is the standard form of 0.000005678.

Question.108 Express the product of 3.2 x 10⁶ and 4.1 x 10¹ in the standard form.
Solution.

Question.109

Solution.

Question. 110 Some migratory birds travel as much as 15000 km to escape the extreme climatic conditions at home. Write the distance in metres using scientific notation.
Solution.

Question. 111 Pluto is 5913000000 m from the Sun. Express this in the standard form.
Solution.

Question. 112 Special balances can weigh something as 0.00000001 gram. Express this number in the standard form.
Solution.

Question. 113 A sugar factory has annual sales of 3 billion 720 million kilograms of sugar. Express this number in the standard form.
Solution.

Question. 114 The number of red blood cells per cubic millimetre of blood is approximately mm³)
Solution. The average body contain 5 L of blood.
Also, the number of red blood cells per cubic millimetre of blood is approximately 5.5 million.
Blood contained by body = 5 L = 5 x 100000 mm³
Red blood cells = 5 x 100000 mm³
Blood = 5.5 x 1000000 x 5 x 100000= 55 x 5 x 10^{5 + 5}
= 275 x 10¹⁰ = 2.75 x 10¹⁰ x 10² = 2.75 x 10¹²

Question. 115 Express each of the following in standard form:

Solution.

Question.116

Solution.

Question.117

Solution.

In questions 118 and 119, find the value of n.
Question.118

Solution.

Question.119

Solution.

Question.120

Solution.

Question.121

Solution.

Question.122

Solution.

Question.123 A new born bear weights 4 kg. How many kilograms might a five year old bear weight if its weight increases by the power of 2 in 5 yr?
Solution.
Weight of new born bear = 4 kg
Weight increases by the power of 2 in 5 yr.
Weight of bear in 5 yr = (4)² = 16 kg

Question.124 The cell of a bacteria doubles in every 30 min. A scientist begins with a single cell. How many cells will be thereafter (a) 12 h (b) 24 h ?
Solution.

Question.125 Planet A is at a distance of 9.35 x 10⁶ km from Earth and planet B is 6.27 x 107 km from Earth. Which planet is nearer to Earth?
Solution.
Distance between planet A and Earth = 9.35 x 10⁶ km Distance between planet B and Earth = 6.27 x 10⁷ km
For finding difference between above two distances, we have to change both in same exponent of 10, i.e. 9.35 x.10⁶ = 0.935 x 10⁷, clearly 6.27 x 10⁷ is greater.
So, planet A is nearer to Earth.

Question.126 The cells of a bacteria double itself every hour. How many cells will be there after 8 h, if initially we start with 1 cell. Express the answer in powers.
Solution.

Question. 127 An insect is on the 0 point of a number line, hopping towards 1. She covers half the distance from her current location to 1 with each hop.
So, she will be at 1/2 after one hop, 3/4 after two hops and so on.

(a) Make a table showing the insect’s Location for the first 10 hops.
(b) Where will the insect be after n hops?
(c) Will the insect ever get to 1? Explain.
Solution.
(a) On the basis of given information in the question, we can arrange the following table which shows the insect’s location for the first 10 hops.

Question. 128 Predicting the ones digit, copy and complete this table and answer the questions that follow.

Solution.
(a) On the basis of given pattern in 1^x and 2^x , we can make more patterns for 3^x 4^x , 5^x ,6^x , 7^x , 8^x , 9^x , 10^x .
Thus, we have following table which shows all details about the patterns.

Question. 129 Astronomy The table shows the mass of the planets, the Sun and the Moon in our solar system.

Solution.

Question. 130 Investigating Solar System The table shows the average distance from each planet in our solar system to the Sun.

Solution.

Question. 131 This table shows the mass of one atom for five chemical elements.
Use it to answer the question given.

(a) Which is the heaviest element?
(b) Which element is lighter, Silver or Titanium?
(c) List all the five elements in order from lightest to heaviest.
Solution.

Question. 132 The planet Uranus is approximately 2,896,819,200,000 metres away from the Sun. What is this distance in standard form?
Solution.
Distance between the planet Uranus and the Sun is 2896819200000 m.
Standard form of 2896819200000 = 28968192 x 10 x 10 x 10 x 10 x 10
= 28968192 x 10⁵ = 2.8968192 x 10¹² m

Question. 133 An inch is approximately equal to 0.02543 metres. Write this distance in standard form.
Solution. Standard form of 0.02543 m = 0.2543 x 10^-1 m = 2.543 x 10^-2 m Hence,’ standard form of 0.025434s 2.543 x 10^-2 m.

Question.134 The volume of the Earth is approximately 7.67 x 10^-7 times the volume
of the Sun. Express this figure in usual form.
Solution.

Question.135 An electron’s mass is approximately 9.1093826 x 10^-31 kilograms. What is its mass in grams?
Solution.

Question. 136 At the end of the 20th century, the world population was approximately 6.1 x 10⁹ people. Express this population in usual form. How would you say this number in words?
Solution.
Given, at the end of the 20th century, the world population was 6,1 x 10⁹ (approx). People population in usual form = 6.1 x 10⁹ = 6100000000 Hence, population in usual form was six thousand one hundred million.

Question.137 While studying her family’s history, Shikha discovers records of ancestors 12 generations back. She wonders how many ancestors she had in the past 12 generations. She starts to make a diagram to help her figure this out. The diagram soon becomes very complex

Solution.
(a) On the basis of given diagram, we can make a table that shows the number of ancestors in each of the 12 generations.

Question. 138 About 230 billion litres of water flows through a river each day, how many litres of water flows through that river in a week? How many litres of water flows through the river in an year? Write your answer in standard notation.
Solution.

Question. 139 A half-life is the amount of time that it takes for a radioactive substance to decay one-half of its original quantity.
Suppose radioactive decay causes 300 grams of a substance to decrease 300 x 2^-3 grams after 3 half-lives. Evaluate 300 x 2^-3to determine how many grams of the substance is left.
Explain why the expression 300 x 2^-n can be used to find the amount of the substance that remains after n half-lives.
Solution.

Question. 140 Consider a quantity of a radioactive substance. The fraction of this quantity that remains after t half-lives can be found by using the expression 3^-t.
(a) What fraction of the substance remains after 7 half-lives?
(b) After how many half-lives will the fraction be 1/243 of the original?
Solution.

Question. 141 One fermi is equal to 10^-15 metre. The radius of a proton is 1.3 fermi. Write the radius of a proton (in metres) in standard form.
Solution. The radius of a proton is 1.3 fermi.
One fermi is equal to 10^-15 m.
So, the radius of the proton is 1.3 x 10^-15 m.
Hence, standard form of radius of the proton is 1.3 x  10^-15 m.

Question. 142 The paper clip below has the indicated length. What is the length in Standard form.

Solution.

Length of the paper clip = 0.05 m
In standard form, 0.05 m = 0.5 x 10^-1 = 5.0 x 10^-2 m
Hence, the length of the paper clip in standard form is 5.0 x 10^-2 m

Question.143 Use the properties of exponents to verify that each statement is true.

Solution.

Question. 144 Fill in the blanks.

Solution.

Question. 145 There are 86400 sec in a day. How many days long is a second? Express your answer in scientific notation.
Solution. Total seconds in a day = 86400
So, a second is long as 1/86400 = 0.000011574
Scientific notation of 0.000011574= 1.1574 x 10^-5days

Question. 146 The given table shows the crop production of a state in the year 2008 and 2009. Observe the table given below and answer the given questions.

(a) For which crop(s) did the production decrease?
(b) Write the production of all the crops in 2009 in their standard form.
(c) Assuming the same decrease in rice production each year as in 2009, how many acres will be harvested in 2015? Write in standard form.
Solution.

Question. 147 Stretching Machine
Suppose you have a stretbhirtg machine which could stretch almost anything, e.g. If you put a 5 m stick into a (x 4) stretching machine (as shown below), you get a 20 m stick.
Now, if you put 10 cm carrot into a (x 4) machine, how long will it be when it comes out?

Solution.
According to the question, if we put a 5m stick into a (x 4) stretching machine, then machine produces 20 m stick.
Similarly, if we put 10 cm carrot into a (x 4) stretching machine, then machine produce 10 x 4= 40 cm stick.

Question. 148 Two machines can be hooked together. When something is sent through this hook up, the output from the first machine becomes the input for the second.

Solution.

Question. 149 Repeater Machine
Similarly, repeater machine is a hypothetical machine which automatically enlarges items several times, e.g. Sending a piece of wire through a (x 2⁴) machine is the same as putting it through a (x 2) machine four times. ‘
So, if you send a 3 cm piece of wire thorugh a (x 2)4 machine, its length becomes 3 x 2 x 2 x 2 x 2 = 48 cm. It can also be written that a base (2) machine is being applied 4 times.

What will be the new length of a 4 cm strip inserted in the machine?
Solution.
According to the question, if we put a 3 cm piece of wire through a (x 2⁴) machine, its length becomes 3 x 2 x 2 x 2 x 2 = 48 cm.
Similarly, 4 cm long strip becomes 4 x 2 x 2 x 2 x 2 = 64 cm.

Question. 150 For the following repeater machines, how many times the base machine is applied and how much the total stretch is?

Solution.
In machine (a), (x 100 ²) = 10000stretch. Since, it is two times the base machine.
In machine (b), (x 7 ⁵) = 16807 stretch.
Since, it is fair times the base machine.
In machine (c), (x 5⁷) = 78125stretch.
Since, it is 7 times the base machine.

Question. 151 Find three repeater machines that will do the same work as a (x 64) machine. Draw them, or describe them using exponents.’
Solution.
We know that, the possible factors of 64 are 2, 4, 8. :
If 2⁶ =64, 4³ =64 and 8² =64
Hence, three repeater machines that would work as a (x 64) will be (x 2⁶ ), (x 4³) and (x 8²). The diagram of (x 2⁶), (x 4³)and (x 8²)is given below.

Question. 152 What will the following machine do to a 2 cm long piece of chalk?

Solution.
The machine produce x 1¹⁰⁰=1
So, if we insert 2 cm long piece of chalk in that machine, the piece of chalk remains same.

Question. 153 In a repeater machine with 0 as an exponent, the base machine is applied 0 times.
(a) What do these machines do to a piece of chalk?

(b) What do you think the value of 6° is?
You have seen that a hookup of repeater machines with the same base can be replaced by a single repeater machine. Similarly, when you multiply exponential expressions with the same base, you can replace them with a single expression.
Asif Raza thought about how he could rewrite the expression 220 x 2⁵.

Asif Raza’s idea is one of the product laws of exponents, which can be expressed like this
Multiplying Expressions with the Same Base ab x ac = ab+ c
Actually, this law can be used with more than two expressions. As long as the bases are the same, to find the product you can add the exponents and use the same base.
Solution.

Question. 154 Shrinking Machine In a shrinking machine, a piece of stick is compressed to reduce its length. If 9 cm long sandwich is put into the shrinking machine below, how long will it be when it emerges?

Solution.
According to the question, in a shrinking machine, a piece of stick is compressed to reduce its length. If 9 cm long sandwich is put into the shrinking machine, then the length
of sandwich will be 9 x 1/ 3^-1= 9 x 3 = 27 cm.

Question. 155 What happens when 1 cm worms are sent through these hook-ups?

Solution.

Question. 156 Sanchay put a 1 cm stick of gum through a (1 x 3^-2) machine. How
long was the stick when it came out?
Solution.

Question. 157 Ajay had a 1 cm piece of gum. He put it through repeater machine
given below and it came out 1/100000 cm long. What is the missing
value?

Solution.

Question. 158 Find a single machine that will do the same job as the given hook-up.


Solution.

Question. 159 Find a single repeater machine that will do the same work as each hook-up.


Solution.

Question. 160 For each hook-up, determine whether there is a single repeater
machine that will do the same work. If so, describe or draw it.

Solution.

Question. 161 Shikha has an order from a golf course designer to put palm trees through a (x 2³) machine and then through a (x 3³) machine. She thinks that she can do the job with a single repeater machine. What single repeater machine she should use?

Solution.
Sol. The work done by hook-up machine is equal to 2 x 2 x 2 x 3 x 3 x 3 = 216 = 6³ So, she should use (x 6³) single machine for the purpose.

Question. 162 Neha needs to stretch some sticks to 25² times of their original lengths, but her (x 25) machine is broken. Find a hook-up of two repeater machines that will do the same work as a (x 25²) machine. To get started, think about the hook-up you could use to replace the (x 25) machine.

Solution.
Work done by single machine (x 25²) = 25 x 25 = 625 or 5 x 5 x 5 x 5 or 52 x 52

Hence, (x 5²) and (x 5²) hook-up machine can replace the (x 25) machine.

Question.163 Supply the missing information for each diagram.

solution.

Question. 164 If possible, find a hook-up of prime base number machine that will do the same work as the given stretching machine. Do not use (x 1) machines.

solution.
(a) Single machine work = 100
Hook-up machine of prime base number that do the same work down by x 100
= 2² x 5²
=4×25
= 100
(b) x 99 = 3² x 111 hook-up machine.
(c) x 37 machine cannot do the same work.
(d) x 1111 = 101 x 11 hook-up machine.

Question. 165 Find two repeater machines that will do the same work as a (x 81) machine.
Solution. Two repeater machines that do the same work as (x 81) are (x 3⁴) and (x 9²).
Since, factor of 81 are.3 and 9.

Question. 166

Solution.

Question. 167 Find three machines that can be replaced with hook-up of (x 5) machines.
Solution.
Since, 5² = 25, 5³ = 125, 5⁴ = 625
Hence, (x 5²), (x 5³)and (x 5⁴) machine can replace (x 5) hook-up machine.

Question. 168 The left column of the chart lists is the length of input pieces of ribbon. Stretching machines are listed across the top.
The other entries are the outputs for sending the input ribbon from that row through the machine from that column. Copy and complete the chart.

Solution.
In the given table, the left column of chart list is the length of input piece of ribbon. Thus, the outputs for sending the input ribbon are given in the following table.

Question. 169 The left column of the chart lists is the length of input chains of gold. Repeater machines are listed across the top. The other entries are the outputs you get when you send the input chain from that row through the repeater machine from that column. Copy and complete the chart.

Solution.
In the given table, the left column of the chart lists is the length of input chains of gold. Thus, the output we get when we send the input chain from the row through the repeater machine are detailed in the following table.

Question.170 Long back in ancient times, a farmer saved the life of a king’s daughter. The king decided to reward the farmer with whatever he wished. The farmer, who was a chess champion, made an unusal request
“I would like you to place 1 rupee on the first square of my chessboard. 2 rupees on the second square, 4 on the third square, 8 on the fourth square and so on, until you have covered all 64 squares. Each square should have twice as many rupees as the previous square.” The king thought this to be too less and asked the farmer to think of some , better reward, but the farmer didn’t agree.
How much money has the farmer earned?
[Hint The following table may help you. What is the first square on which the king will place atleast Rs. 10 lakh?]

Solution.

Question. 171 The diameter of the Sun is 1.4 x 10⁹ m and the diameter of the Earth is 1.2756 x 10⁷ m. Compare their diameters by division.
Solution.

Question. 172 Mass of Mars is 6.42 x 10²⁹ kg and mass of the Sun is 1.99 x 10³⁰ kg. What is the total mass?
Solution.
Mass of Mars = 6.42 x 10²⁹  kg
Mass of the Sun = 1.99 x 10³⁰ kg
Total mass of Mars and Sun together = 6.42 x 10²⁹  + 1.99 x 10³⁰
= 6.42 x 10²⁹  + 19.9 x 10²⁹  = 26.32 x 10²⁹  kg

Question. 173 The distance between the Sun and the Earth is 1.496 x 10⁸ km and : distance between the Earth and the Moon is 3.84 x 10⁸ m. During
solar eclipse, the Moon comes in between the Earth and the Sun. What is the distance between the Moon and the Sun at that particular time?
Solution.
The distance between the Sun and the Earth is 1.496 x 10s km
= 1.496 x 10⁸ x 10³ m = 1496 x 10⁸ m
The distance between the Earth and the Moon is 3.84 x10⁸ m.
The distance between the Moon and the Sun at particular time (solar eclipse) = (1496 x 10⁸-3.84 x 10⁸)m = 1492.   16 x 10⁸ m

Question. 174 A particular star is at a distance of about 8.1 x 10¹³ km from the Earth. Assuring that light travels at 3 x 10⁸ m per second, find how long does light takes from that star to reach the Earth?
Solution.

Question. 175 By what number should  (-5)^-1 be divided so that the quotient may be equal to  (-5)^-1?
Solution.

Question. 176 By what number should (-8)^-3 .be multiplied so that the product may be equal to  (-6)^-3?
Solution.

Question. 177 Find x.

Solution.

Question. 178 If a = – 1, b = 2,-then find the value of the following,
(i) a^b + b^a   (ii) a^b  – b^a
(iii) a^b  x b^a (iv) a^{b ÷} b^a
Solution.

Question.179 Express each of the following in exponential form.

Solution.

Question. 180 Simplify

Solution.

The post NCERT Exemplar Problems Class 8 Mathematics Exponents and Powers appeared first on Learn CBSE.